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Simplifying Trigonometric IdentitiesSimplifying Trigonometric Identities
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Question 1 of 7
1. Question
Which of the following are trigonometric identities?Hint
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Trigonometric Identities
$$\sec\theta=\frac{1}{\text{cos}\theta}$$$$\csc\theta=\frac{1}{\text{sin}\theta}$$$$\cot\theta=\frac{1}{\text{tan}\theta}$$$$\tan\theta=\frac{\text{sin}\theta}{\text{cos}\theta}$$$$\cot\theta=\frac{\text{cos}\theta}{\text{sin}\theta}$$Using the given image, we can see that the radius of the circle will be `1`. Therefore, the circle follows the formula `x^2+y^2=1`, where `1` is the radius.Make a reference triangle by making a line from the point of origin `(0,0)` going to any point in the circle to represent the radius and connect it to either the x or y-axis
Notice that, given angle `theta`, the circle has the vertical line parallel to the y-axis as its opposite side and the line on the x-axis as the adjacent sideNow that we have the opposite and adjacent sides and the radius `1` as the hypotenuse, we can use the trigonometric functions to find their respective values`\text(sin)` `=` `\text(opposite)/(\text(hypotenuse))` `=` `y/1` `=` `y` `\text(cos)` `=` `\text(adjacent)/(\text(hypotenuse))` `=` `x/1` `=` `x` `x` and `y` are both coordinate values, so we can write the coordinates as `(\text(cos) theta,\text(sin) theta)`Finally, use the values of `x` and `y` and derive its trigonometric identities`x^2+y^2` `=` `1` `\text(cos)^2theta+\text(sin)^2theta` `=` `1` Substitute known values `(\text(cos)^2theta+\text(sin)^2theta)``divide \text(cos)^2theta` `=` `1``divide \text(cos)^2theta` Divide both sides by `\text(cos)^2theta` `1+(\text(sin)^2theta)/(\text(cos)^2theta)` `=` `1/(\text(cos)^2theta)` `1+``\text(tan)^2theta` `=` `1/(\text(cos)^2theta)` `(\text(sin)^2theta)/(\text(cos)^2theta)=\text(tan)^2theta` `1+\text(tan)^2theta` `=` `\text(sec)^2theta` `1/(\text(cos)^2theta)=\text(sec)^2theta` `\text(sec)^2theta` `=` `1+\text(tan)^2theta` `x^2+y^2` `=` `1` `\text(cos)^2theta+\text(sin)^2theta` `=` `1` Substitute known values `(\text(cos)^2theta+\text(sin)^2theta)``divide \text(sin)^2theta` `=` `1``divide \text(sin)^2theta` Divide both sides by `\text(sin)^2theta` `(\text(cos)^2theta)/(\text(sin)^2theta)+1` `=` `1/(\text(sin)^2theta)` `\text(cot)^2theta``+1` `=` `1/(\text(sin)^2theta)` `(\text(cos)^2theta)/(\text(sin)^2theta)=\text(cot)^2theta` `\text(cot)^2theta+1` `=` `\text(csc)^2theta` `1/(\text(sin)^2theta)=\text(csc)^2theta` `\text(csc)^2theta` `=` `\text(cot)^2theta+1` `\text(sec)^2theta=1+\text(tan)^2theta``\text(csc)^2theta=\text(cot)^2theta+1` -
Question 2 of 7
2. Question
Simplify`1/(1+sin theta)+1/(1-sin theta)`Hint
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Trigonometric Identities
$$\sec\theta=\frac{1}{\text{cos}\theta}$$$$\csc\theta=\frac{1}{\text{sin}\theta}$$$$\cot\theta=\frac{1}{\text{tan}\theta}$$$$\tan\theta=\frac{\text{sin}\theta}{\text{cos}\theta}$$$$\cot\theta=\frac{\text{cos}\theta}{\text{sin}\theta}$$Apply the rule of adding fractions and simplify the expression`1/(1+\text(sin) theta)+1/(1-\text(sin) theta)` `=` `(1-\text(sin) theta+1+\text(sin) theta)/((1+\text(sin) theta)(1-\text(sin) theta))` `=` `2/((1+\text(sin) theta)(1-\text(sin) theta))` Combine like terms `=` `2/(1-\text(sin)^2theta)` Expand Recall that `\text(sin)^2theta+\text(cos)^2theta=1`. Derive this formula to further simplify the expression`\text(sin)^2theta+\text(cos)^2theta` `=` `1` `\text(sin)^2theta+\text(cos)^2theta` `-\text(sin)^2theta` `=` `1` `-\text(sin)^2theta` Subtract `\text(sin)^2theta` from both sides `\text(cos)^2theta` `=` `1-\text(sin)^2theta` `2/(1-\text(sin)^2theta)` `=` `2/(\text(cos)^2theta)` `1-\text(sin)^2theta=\text(cos)^2theta` `=` `2/1xx 1/(\text(cos)^2theta)` Take out `2` `=` `2xx\text(sec)^2theta` `1/(\text(cos)^2theta)=\text(sec)^2theta` `=` `2\text(sec)^2theta` `2\text(sec)^2theta` -
Question 3 of 7
3. Question
Simplify`(tan theta)/(sec theta)`Hint
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Trigonometric Identities
$$\sec\theta=\frac{1}{\text{cos}\theta}$$$$\csc\theta=\frac{1}{\text{sin}\theta}$$$$\cot\theta=\frac{1}{\text{tan}\theta}$$$$\tan\theta=\frac{\text{sin}\theta}{\text{cos}\theta}$$$$\cot\theta=\frac{\text{cos}\theta}{\text{sin}\theta}$$Substitute known values to `\text(tan)` and `\text(sec)``\text(tan)` `=` `(\text(sin))/(\text(cos))` `\text(sec)` `=` `1/(\text(cos))` `(\text(tan) theta)/(\text(sec) theta)` `=` `((\text(sin))/(\text(cos)))/(1/(\text(cos)))` Substitute known values `=` `\text(sin) theta` Apply rule of dividing fractions `\text(sin) theta` -
Question 4 of 7
4. Question
Simplify`sec theta-sec theta sin^2theta`Hint
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Trigonometric Identities
$$\sec\theta=\frac{1}{\text{cos}\theta}$$$$\csc\theta=\frac{1}{\text{sin}\theta}$$$$\cot\theta=\frac{1}{\text{tan}\theta}$$$$\tan\theta=\frac{\text{sin}\theta}{\text{cos}\theta}$$$$\cot\theta=\frac{\text{cos}\theta}{\text{sin}\theta}$$Factor out `\text(sec) theta` on both terms`\text(sec) theta-\text(sec) theta \text(sin)^2theta` `=` `\text(sec) theta(1-\text(sin)^2theta)` Factorise Recall that `\text(sin)^2theta+\text(cos)^2theta=1`. Derive this formula to further simplify the expression`\text(sin)^2theta+\text(cos)^2theta` `=` `1` `\text(sin)^2theta+\text(cos)^2theta` `-\text(sin)^2theta` `=` `1` `-\text(sin)^2theta` Subtract `\text(sin)^2theta` from both sides `\text(cos)^2theta` `=` `1-\text(sin)^2theta` `\text(sec) theta(1-\text(sin)^2theta)` `=` `\text(sec) theta(\text(cos)^2theta)` `1-\text(sin)^2theta=\text(cos)^2theta` `=` `1/(\text(cos) theta)*(\text(cos)^2theta)` `\text(sec) theta=1/(\text(cos) theta)` `=` `(\text(cos)^2theta)/(\text(cos) theta)` `=` `\text(cos) theta` `\text(cos) theta` -
Question 5 of 7
5. Question
Simplify`sec^2A-csc^2A`Hint
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The difference of two squares is a format of a quadratic expression where there are two terms that are both perfect squares.First, find the square of the two terms.First term`\text(sec)^2A` `=` `\text(sec) A*\text(sec) A` Second term`\text(csc)^2A` `=` `\text(csc) A*\text(csc) A` Finally, get the sum and difference of the two values.`(``\text(sec) A``-``\text(csc) A``)(``\text(sec) A``+``\text(csc) A``)` `(\text(sec) A-\text(csc) A)(\text(sec) A+\text(csc) A)` -
Question 6 of 7
6. Question
Simplify`(tan^2theta)/(sec theta-1)`Hint
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Trigonometric Identities
$$\sec\theta=\frac{1}{\text{cos}\theta}$$$$\csc\theta=\frac{1}{\text{sin}\theta}$$$$\cot\theta=\frac{1}{\text{tan}\theta}$$$$\tan\theta=\frac{\text{sin}\theta}{\text{cos}\theta}$$$$\cot\theta=\frac{\text{cos}\theta}{\text{sin}\theta}$$Recall that `1+\text(tan)^2theta=\text(sec)^2theta`. Derive this formula to further simplify the expression`1+\text(tan)^2theta` `=` `\text(sec)^2theta` `1+\text(tan)^2theta` `-1` `=` `\text(sec)^2theta` `-1` Subtract `1` from both sides `\text(tan)^2theta` `=` `\text(sec)^2theta-1` `(\text(tan)^2theta)/(\text(sec) theta-1)` `=` `(\text(sec)^2theta-1)/(\text(sec) theta-1)` `\text(tan)^2theta=\text(sec)^2theta-1` `=` `((\text(sec)theta-1)(\text(sec)theta+1))/(\text(sec) theta-1)` Factorise `=` `\text(sec)theta+1` `(\text(sec) theta-1)/(\text(sec) theta-1)=1` `\text(sec)theta+1` -
Question 7 of 7
7. Question
Simplify`((tan^2beta)(cot^2beta+1))/(tan^2beta+1)`- (1)
Hint
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Trigonometric Identities
$$\sec\theta=\frac{1}{\text{cos}\theta}$$$$\csc\theta=\frac{1}{\text{sin}\theta}$$$$\cot\theta=\frac{1}{\text{tan}\theta}$$$$\tan\theta=\frac{\text{sin}\theta}{\text{cos}\theta}$$$$\cot\theta=\frac{\text{cos}\theta}{\text{sin}\theta}$$Recall that `1+\text(cot)^2theta=\text(csc)^2theta`. Derive this formula to further simplify the expression`((\text(tan)^2beta)(\text(cot)^2beta+1))/(\text(tan)^2beta+1)` `=` `(\text(tan)^2beta*\text(csc)^2theta)/(\text(tan)^2beta+1)` `1+\text(cot)^2theta=\text(csc)^2theta` Finally, substitute known values to `\text(tan)`, `\text(csc)`, and `\text(sec)``\text(tan)` `=` `(\text(sin))/(\text(cos))` `\text(csc)` `=` `1/(\text(sin))` `\text(sec)` `=` `1/(\text(cos))` `(\text(tan)^2beta*\text(csc)^2theta)/(\text(tan)^2beta+1)` `=` `((\text(sin)^2beta)/(\text(cos)^2beta)*1/(\text(csc)^2theta))/(1/(\text(cos)^2theta))` Substitute known values `=` `(\text(sin)^2beta)/(\text(cos)^2beta)*1/(\text(csc)^2theta)*(\text(cos)^2theta)/1` Apply rule of dividing fractions `=` `1` `1`